# Orthogonal group representation induces an isometry on $L^p$ spaces

I am reading a paper and I am unsure about the following. I will first explain the setup.

Suppose we have a measure space $(X,B,\mu)$. Let $G$ be a compact Lie group with Haar measure $\nu$. Suppose that $\mathbb{R}^d$ is a representation of $G$, and that the image of $G\to GL_d$ lies in the orthogonal group $O_d$. Let $h:X\to G$ be measurable, and define $M_h v(x)=h(x)v(x)$ where $v\in L^p(X,\mathbb{R}^d)$ for $p\geq 1$. Apparently, $M_h$ is an isometry on $L^p(X,\mathbb{R}^d)$.

I believe the answer should use the fact that the representation lies in the orthogonal group, but I am not so familiar with representation theory in general. Could somebody please help?

This isn't really about group representations. If $A$ is an orthogonal matrix, then it preserves Euclidean norm: $\|Av\|=\|v\|$ for every $v\in\mathbb{R}^d$. Consequently, $\|h(x)v(x)\| = \|v(x)\|$ for all $x\in X$, which implies $$\|M_h v\|_{L^p} = \left(\int_X \|h(x)v(x)\|^p \,d\mu(x) \right)^{1/p} = \left(\int_X \|v(x)\|^p \,d\mu(x) \right)^{1/p} = \|v\|_{L^p}$$